Question

An object 6 cm in length is held 24 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer 1

Explanation:

Edit the following passage

In India,teacher have been accorded the (a)___,___

highest importance. India has been blessed of (b)_

__,___

a long line of eminent teacher

(c)___,___. Their teaching have not been (d)___,___

confined in the classrooms (e)___,___

and has been extended to the whole world (f)___,___.Answer fast, I will mark the first answer as brainiest

Answer 2

Explanation:

$\huge\bf\underline{\red{❥A}\purple{n}\pink{s}\orange{w}\blue{e}\green{r}}$

Position of object, u = -24 cm

Focal length of the lens, f = 10 cm

Position of image, v = ?

We know that (WKT),

$\frac{1}{v} \: - \: \frac{1}{u} = \frac{1}{f} \\ \\ \frac{1}{v} \: - \: \frac{1}{24} = \frac{1}{10} \\ \\ \frac{1}{v} = \frac{1}{10} \: - \: \frac{1}{24}$

___________________________________

So,

$\frac{1}{v} = \frac{(6 \: - \: 2)}{120}$

____________

That is,

$\frac{1}{v} = \frac{4}{120}$

____________

So,

$v = \frac{120}{4} = \: 30 \: cm$

____________

Thus,

$m \: = \: \frac{30}{-24} = 1.25$

___________________________________

Also,

$m \: = \frac{height \: of \: image}{height \: of \: object} \\ \\ - 1.25 \: = \: \frac{height \: of \: image}{6 \: cm} \\ \\ height \: of \: image = 1.25 \: \times \: 6 \\ \\ = 7.50 \: or \: 7.5$

_______________________________

Therefore, height of the image = 7.5 cm

The negative sign of the height of image shows that an inverted image is formed.

Thus, the position of image at 30 cm on opposite side of lens.

Size of image = 7.5 cm at the opposite side of length.

Nature of image is real and inverted.